Let us work over $\mathbb C$. In this article by S. Katz, it is stated that for a quintic threefold $X\subset \mathbb P^4$ one has $$H_2(X,\mathbb Z)\cong\mathbb Z.$$ Can anyone help me to see why this is true?

Remark. If $X$ was a general surface, this would be easier, as we would have $$H_2(X,\mathbb Z)\cong H^2(X,\mathbb Z)=\textrm{Pic}(X)=\mathbb Z.$$ The last equality uses generality (I guess this is really necessary, but I lack a counterexample). That is why I would also accept an answer assuming the threefold $X$ to be general. However, in the quoted paper the result is stated with no genericity assumption, so it should be true as stated.

The group in question, $H_2(X,\mathbb Z)$, is now isomorphic to $H^4(X,\mathbb Z)$. I do not find it easier to work with the latter, though.

I do not know if I can use the exponential sequence, as taking its cohomology would give me a sequence of vector spaces, and the group $H^4(X,\mathbb Z)$ does not even appear there. Any hint or solution would be much appreciated.

I was also wondering whether $H_2(X,\mathbb Z)\cong\mathbb Z$ can hold for other projective Calabi-Yau threefolds, different from the quintic in $\mathbb P^4$.

Thank you!

  • 1
    $\begingroup$ mathoverflow.net/a/60331/450 $\endgroup$ Mar 16, 2014 at 19:10
  • $\begingroup$ Dear @GeorgesElencwajg, just found some minutes ago, and upvoted. Great answer! It seems that being CY has nothing to do with the statement, though. $\endgroup$
    – Brenin
    Mar 16, 2014 at 19:16
  • $\begingroup$ Dear @Brenin, thanks for the kind words. You are quite right that Lefschetz's topological result does not need the Calabi-Yau hypothesis. As a confirmation, Lefschetz stated his theorem in 1924, Calabi was born in 1923 and Yau in 1949 :-) $\endgroup$ Mar 16, 2014 at 20:05

1 Answer 1


One simple and powerful answer: Lefschetz Hyperplane Theorem.


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